Find $\lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{4x^{2}+10}}{x-5}$.

Figure 55 $f( x) =\dfrac{\sqrt{4x^{2}+10}}{x-5}$

Solution Divide each term of the numerator and the denominator by $x$, the term with the highest power of $x$ that appears in the denominator. But remember, since $\sqrt{x^{2}}=\vert x\vert =x$ when $x\geq 0$, in the numerator the divisor in the square root will be $x^{2}$. $$\begin{eqnarray*} \lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{4x^{2}+10}}{x-5} &=&\lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{\dfrac{4x^{2}+10}{x^{2}} }}{\dfrac{x-5}{x}}=\lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{\dfrac{ 4x^{2}}{x^{2}}+\dfrac{10}{x^{2}}}}{1-\dfrac{5}{x}}= \dfrac{\lim\limits_{x\rightarrow \infty }\sqrt{4+ \dfrac{10}{x^{2}}}}{\lim\limits_{x\rightarrow \infty }\left[ 1-\dfrac{5}{x} \right] } \\ &=&{\sqrt{\lim\limits_{x\rightarrow \infty }\left[ 4+\dfrac{10}{x^{2}}\right] }}={\sqrt{4}} =2 \end{eqnarray*}$$